OS, GATE2018

Question

Comments

Is ans option B? 8GB?

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LAS : 64 bits

PAS:64 bits (processor)

LAS: 

p1

p2

offset

32                                    16                                     16

p1: no of pages in outer level page table

p2:no of pages in inner level page table

PAS: 

No of frames

offset

48                                                                 16

Page table size=no of pages*PTE.

Every PTE definitely contains frame numbers.

If we PTE= no of frames then

page table size=2^32 * 48 bits

                         =2^32*6B

                         =24GB.

Hence the outer level page table size=24GB. 

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The answer given on ace test solutions is 32GB.

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i think they consider 6B to round off 8B then u get 32GB.

But that is not the actual size of page table.

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I am not getting how PAS is 64 bit.

i mean 64-bit processor means the max data type size it can handle. How is this equal to PAS.

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@santhoshdevulapally How did you calculate the PAS, when it is not given in the question?

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A 64-bit processor. -> every 8 bytes are physically addressed.

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Please help me out in this.

Our page offset bits = 16( in this question). And we also know that the outermost page table should fit in one frame. So in worst case we can have 2^16 addressable locations in a page, so how come the outermost page table is indexed by 32 bits (which implies that the page of outermost page table has 2^32 addressable locations).

Thanks in advance.

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quite a ambiguous question.there has to be way to find PTE size 

by the way u can read arjun sir comments on this question here

https://gateoverflow.in/72799/ace-test-series-os

Answer 1

The total entries possible in outer page table are 2^32 each one of size 8 bytes.. i.e, 2^32 * 8 bytes

4 GB * 8 = 32 GB

Answer = option D.

Comments

How the entry size is 8 bytes ?

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I think the answer is 8GB...2^32 * 2=8GB